Math, asked by belongjkonyak, 3 months ago

show that if the diagonal of a quadrilateral are equal and bisects each other at right angle then it is a square.​

Answers

Answered by mathdude500
0

Given :-

  • Let ABCD is a quadrilateral in which diagonals AC and BD intersects at O.

  • AC =BD,

  • OA = OC,

  • OB = OD,

  • ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°.

To Prove :-

  • ABCD is a square.

Proof :-

Consider,

\rm :\longmapsto\:In \: \triangle \: AOB \: and \:\triangle \: BOC

\rm :\longmapsto\:OA = OC \:  \:  \:  \: (given)

\rm :\longmapsto\:OB = OB \:  \:  \:  \: (common)

\rm :\longmapsto \: \angle \: AOB \:  =  \:\angle \: BOC \:  \:  \: (each \: 90 \degree)

\rm :\implies\: \: \triangle \: AOB \: \cong \:\triangle \: BOC \:  \:  \: (SAS)

\bf\implies \:AB = BC \:  \:  \:  \:  \:  \: (CPCT) -  - (1)

Similarly,

\bf\implies \:BC = CD \:  \:  \:  \:  \ -  -  -  - (2)

Similarly,

\bf\implies \:CD = DA \:  \:  \:  \:  \ -  -  -  - (3)

From equation (1), (2) and (3), we concluded that

\rm :\longmapsto\:AB = BC = CD = DA

Since, opposite pair of sides are equal.

\bf\implies \:ABCD \: is \: a \: parallelogram.

Now,

Consider,

\rm :\longmapsto\:In \: \triangle \: ABD \: and \:\triangle \: ABC

\rm :\longmapsto\:AC = BD \:  \:  \:  \: (given)

\rm :\longmapsto\:AB = AB\:  \:  \:  \: (common)

\rm :\longmapsto\:AD = BC\:  \:  \:  \: (proved \: above)

\rm :\implies\: \: \triangle \: ABD \: \cong \:\triangle \: BAC \:  \:  \: (SSS)

\bf\implies \: \angle \: BAD \:  =  \:  \angle \: ABC \:  \:  \:  \:  \: (CPCT)

Now,

ABCD is a parallelogram,

\bf\implies \: \angle \: BAD  + \angle \: ABC  = 180 \degree  \:  \: (sum \: of \: cointerior)

\bf\implies \: 2\angle \: BAD    = 180 \degree  \:

\bf\implies \: \angle \: BAD    = 90 \degree  \:

Since,

  • ABCD is a parallelogram and ∠BAD = 90°.

Therefore,

  • ABCD is a square.

{\boxed{\boxed{\bf{Hence, Proved}}}}

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